Math 373 - hw 4 solutions
3.2, 3.4, 3.6, 3.8, 3.12, 3.14, 3.20, 3.24;
Tues, 28 Sept 2010
3.2 Let n ∈ N. Prove that if |n − 1| + |n + 1| ≤ 1, then |n2 − 1| ≤ 4.
Since n ∈ N, we have n ≥ 1 and so |n + 1| = n + 1 ≥ 2; hence
|n − 1| + |n + 1| ≥ |n + 1| ≥ 2.
Thus, |n − 1| + |n + 1| 6≤ 1 and therefore the desired result follows vacuously (since the premise is false for all
n ∈ N).
3.4 Let x ∈ R. Prove that if x3 − 5x − 1 ≥ 0, then (x − 1)(x − 3) ≥ −2.
We will use a trivial proof, that is, we will show that (x − 1)(x − 3) ≥ −2 for every x ∈ R. Let x ∈ R be given.
Then since (x − 2)2 ≥ 0 we have by expanding and then completing the square
(x − 1)(x − 3) = x2 − 4x + 3 = (x − 2)2 − 1 ≥ 0 − 1
> −2.
Therefore, (x − 1)(x − 3) ≥ −2 for every x ∈ R; hence, the desired result holds.
3.6 Prove that if x is an odd integer, then 9x + 5 is even.
Let x ∈ Z and suppose that x is odd. Then x = 2k + 1 for some k ∈ Z. Hence,
9x + 5 = 9(2k + 1) + 5 = 18k + 14 = 2(9k + 7).
Since m = 9k + 7 is an integer, 9x + 5 = 2m is even.
3.8 Prove that if a and c are odd integers, then ab + bc is even.
Let a, b, c ∈ Z and suppose that a and c are odd integers. Then a = 2j + 1 and c = 2k + 1 for some j, k ∈ Z. Then
ab + bc = (2j + 1)b + b(2k + 1) = b(2j + 2k + 2)
= 2b(j + k + 1).
Since m = b(j + k + 1) is an integer, ab + bc = 2m is even.
3.12 Let x ∈ Z. Prove that if 7x + 5 is odd, then x is even.
We prove this using the the contrapositive. Suppose that x is odd. Then x = 2k + 1 for some k ∈ Z. Hence,
7x + 5 = 7(2k + 1) + 5 = 14k + 12 = 2(7k + 6).
Since m = 7k + 6 is an integer, 7x + 5 = 2m is even.
3.14 Let x ∈ Z. Prove that 5x − 11 is even if and only if x is odd.
We first prove the forward implicaton by contrapositive. Suppose that x is even. Then x = 2k for some k ∈ Z.
Hence,
5x − 11 = 5(2k) − 11 = 10k − 11 = 2(5k − 6) + 1.
Since m = 5k − 6 is an integer, 5x − 11 = 2m + 1 is odd.
For the converse, suppose x is odd. Then x = 2k + 1 for some k ∈ Z. Hence,
5x − 11 = 5(2k + 1) − 11 = 10k + 5 − 11 = 2(5k − 3).
Since m = 5k − 3 is an integer, 5x − 11 = 2m is even.
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3.20 Prove that if n ∈ Z, then n − n is even.
Let n ∈ Z. There are two cases to consider: either n is even or n is odd.
(i) Suppose first that n is even. Then n = 2k for some k ∈ Z. Hence,
n3 − n = n(n2 − 1) = 2k(4k 2 − 1).
Since m = k(4k 2 − 1) is an integer, n3 − n = 2m is even.
(ii) Suppose next that n is odd. Then n = 2k + 1 for some k ∈ Z.
n3 − n = n(n2 − 1) = (2k + 1)(4k 2 + 4k + 1 − 1)
= 4k(k + 1)(2k + 1).
Since m = 2k(k + 1)(2k + 1) is an integer, n3 − n = 2m is even.
In both cases, n3 − n is even and so the desired result follows.
3.24 Let a, b ∈ Z. Prove that if a + b and ab are of the same parity, then a and b are even.
We prove this using the the contrapositive. Suppose that a is odd or b is odd. There are two cases to consider:
either both are odd or one the two is odd.
(i) Suppose first that both a and b are odd. Then a = 2j + 1 and b = 2k + 1 for some j, k ∈ Z. Hence,
a + b = 2j + 1 + 2k + 1 = 2(j + k + 1)
ab = (2j + 1)(2k + 1) = 2(2jk + j + k) + 1.
Since m = j + k + 1 and n = 2jk + j + k are both integers, a + b = 2m is even and ab = 2n + 1 is odd. Hence,
a + b and ab have opposite parity
(ii) Now suppose that one of two is odd, that is, a and b have opposite parity. Without loss of generality we may
assume that a is odd and b is even. Then a = 2j + 1 and b = 2k for some j, k ∈ Z. Hence,
a + b = 2j + 1 + 2k = 2(j + k) + 1
ab = (2j + 1)(2k) = 2k(2j + 1).
Since m = j + k and n = k(2j + 1) are both integers, a + b = 2m + 1 is odd and ab = 2n + 1 is even. Hence,
a + b and ab have opposite parity.
Thus the desired result follows.
It is also possible to prove the result directly but there are now three cases to consider:
Case (i) a and b are both even, case (ii) a and b are both odd, case (iiii) a and b have opposite parity.
(i) Suppose a and b are both even. In this case the conclusion of the implication is true so the result for this
case follows trivially.
(ii) Suppose a and b are both odd. The result for this case will follow vacuously if we can show that a + b and
ab have opposite parity. But this was proved in case (i) in the first proof above.
(iii) Now suppose that a and b have opposite parity. Then arguing as in case (ii) in the first proof we see that
a + b and ab also have opposite parity. Hence our result holds vacuously in this case also.